1996 AL Winning Percentage by Team
Every team's total, added up from its players' lines. Wins per decision
The Cleveland Indians led the AL in winning percentage in 1996, with .615, .047 ahead of the New York Yankees. The average across 14 AL teams is .500.
Led the AL in 1996
Cleveland Indians
161 games
.615
Winning Percentage
.047 clear of New York Yankees
| # | Team | G | W-L% |
|---|---|---|---|
| 1 |
Cleveland Indians
|
161 | .615 |
| 2 |
 New York Yankees
|
162 | .568 |
| 3 |
 Texas Rangers
|
163 | .556 |
| 4 |
 Baltimore Orioles
|
163 | .543 |
| 5 |
 Seattle Mariners
|
161 | .528 |
| 6 |
 Boston Red Sox
|
162 | .525 |
| 6 |
 Chicago White Sox
|
162 | .525 |
| AL average | .500 | ||
| 8 |
 Milwaukee Brewers
|
162 | .494 |
| 9 |
 Minnesota Twins
|
162 | .482 |
| 9 |
 Oakland Athletics
|
162 | .482 |
| 11 |
 Kansas City Royals
|
161 | .466 |
| 12 |
 Toronto Blue Jays
|
162 | .457 |
| 13 |
 California Angels
|
161 | .435 |
| 14 |
 Detroit Tigers
|
162 | .327 |
A team's total is its players' numbers for the club added up, so a player traded mid-season counts for each team only what he did there. Rates such as ERA and batting average are worked out from the team's summed totals, not averaged across players.